M ATH 234 - E XAM #3
W INTER Q UARTER 2012
Name:
Any student found engaging in academic misconduct will receive a score of 0 on this exam. All suspicious
behavior will be reported to the student misconduct board. In such an instance, you will be force to meet in
front of a board of professors to explain your actions. By signing the space below, you agree that you will
not give nor receive any unauthorized aid during the exam.
Signature:
Page:
3
4
5
6
7
8
9
10
11
Total
Points:
15
10
10
12
5
10
10
13
15
100
Score:
Page:
12
Total
Bonus Points:
4
4
Score:
Final Score:
out of 100 points =
%
- Check that your exam has a total of 14 pages and 13 problems.
- You are allowed to use a calculator on this exam.
- You should show/explain your work on all problems to receive full credit. The correct answer with no
supporting work may result in no credit.
- If you have a question, please come up to the front.
- There is extra paper available up front for scratch work, or if you run out of space.
- Clearly indicate your final answer by placing a box around it.
- You will have 85 minutes to complete your exam.
GOOD LUCK!!!
Math 234
Exam # 3
Page 2 of 14
Math 234
Exam # 3
Page 3 of 14
T RUE OR FALSE .
Answer the following questions True or False. You must justify your answer to receive full credit.
1. (5 points) True/False. Consider the system of n first order equations given by
!u! = P(t)!u + !g(t),
! (t) is a solution to
where P(t) is an n × n matrix of continuous functions (defined for all t). If !u(t) = w
! (t) is also a solution.
the differential equation, then !u(t) = 4 w
2. (5 points) True/False. The functions
!f (1) (t) =
!
e2t
2e2t
"
! 2t "
!f (2) (t) = 4e2t
8e
form a fundamental solution set to the system of differential equations given by
f1! (t)
f2! (t)
=
=
2f1 (t) + 3f2 (t)
f2 (t)
3. (5 points) True/False. If λ is an eigenvalue of the matrix A, then the matrix A − λI is invertible.
Points earned:
out of a possible 15 points
Math 234
Exam # 3
Page 4 of 14
4. (5 points) True/False. Consider the following initial value problem
! "
2
!
!u = A!u,
!u(0) =
.
−2
where A is a constant 2 × 2 matrix with real entries and real eigenvalues λ1 > 0 and λ2 < 0 with
corresponding eigenvectors
! "
! "
1
1
!v(1) =
,
and !v(2) =
.
1
−1
Let !
u(t) represent the solution to the initial value problem. Then the following limit must be true:
lim !u(t) = !0.
t→∞
S HORT A NSWER
5. (5 points) Short Answer. Consider the system of differential equations given by
x!
y
!!
= 3x + 2y
= 3x + 2y + 3y !
If possible, write the above set of differential equations as a system of first order differential equations.
If it is not possible, explain why.
Points earned:
out of a possible 10 points
Math 234
Exam # 3
Page 5 of 14
6. (5 points) Short Answer. If possible, determine a value of β such that all solutions to the system of
differential equation
!
"
1 β
!z,
!z! =
β 1
! "
0
satisfy lim !z(t) =
. If it is not possible to find such a value, explain why.
0
t→∞
7. (5 points) Short Answer. Consider the second order nonlinear initial value problem
y !! + yy ! = t sin(y),
y(0) = y0 .
Outline a method in which you could extend Euler’s numerical method to approximate the solution at
a future time t = h.
Points earned:
out of a possible 10 points
Math 234
Exam # 3
Page 6 of 14
M ATCHING
8. (12 points) Matching. Match the following differential equations with their parametric plots. You
must briefly justify your choice to receive credit.
!
"
0
!y
2
(a)
1
!y =
3
(b)
"
1 1
!y
!y =
−1 1
(c)
3
!y =
5
!
Corresponds to Figure #:
!
!
!
!
Corresponds to Figure #:
"
−1
!y
−3
Corresponds to Figure #:
Figure #2
5
5
4
4
3
3
2
2
1
1
2
0
y
y2
Figure #1
0
ï1
ï1
ï2
ï2
ï3
ï3
ï4
ï4
ï5
ï5
ï5
ï4
ï3
ï2
ï1
0
y1
1
2
3
4
5
ï5
ï4
ï3
ï2
0
y1
1
2
3
4
5
2
3
4
5
Figure #4
5
5
4
4
3
3
2
2
1
1
y2
y2
Figure #3
ï1
0
0
ï1
ï1
ï2
ï2
ï3
ï3
ï4
ï4
ï5
ï5
ï5
ï4
ï3
ï2
ï1
0
y1
1
2
3
4
5
ï5
Points earned:
ï4
ï3
ï2
ï1
0
y1
1
out of a possible 12 points
Math 234
Exam # 3
Page 7 of 14
S OLVE THE F OLLOWING E QUATIONS
9. (5 points) Consider the initial value problem
y ! (t) =
y + 2t2 y 2
,
3
y(.1) = .01
Use one step of the Improved Euler Method to approximate y(.2). Show your work.
Points earned:
out of a possible 5 points
Math 234
Exam # 3
Page 8 of 14
10. Consider the initial value problem
"
−2 2
!,
! =
p
p
−2 −2
!
!
! "
1
!p(0) =
.
2
(a) (10 points) Solve the initial value problem.
Points earned:
out of a possible 10 points
Math 234
Exam # 3
Page 9 of 14
(b) (5 points) Sketch the component plots of the solution to the initial value problem. Be sure to
label your axes.
(c) (5 points) Sketch various solutions to the initial value problem in the phase-plane (parametric
plots). Be sure to label your axes.
Points earned:
out of a possible 10 points
Math 234
Exam # 3
Page 10 of 14
11. (13 points) Find the fundamental solution set to the system


1 2 4
!q! = 0 −2 4 !q.
0 0 2
Points earned:
out of a possible 13 points
Math 234
Exam # 3
Page 11 of 14
12. (15 points) Find the solution to the initial value problem given by
! 4t "
!
"
! "
e
2 −1
0
!
!u + 2t , !u(0) =
!u =
1 4
e
0
Points earned:
out of a possible 15 points
Math 234
13. Bonus: Consider the system
Exam # 3
t !x! = A !x,
Page 12 of 14
t > 0,
where A is a constant 2 × 2 matrix with real entries.
(a) (2 points (bonus)) The system t!x! = A!x is analogous to the second-order Euler equation. Assuming
that !x = !s tr , what is the relationship between r, !s, and A in order to obtain nontrivial solutions
to the differential equation?
(b) (2 points (bonus)) Use your condition from above to find the general solution to the differential
equation
!
"
2 3
!x
t !x! =
0 −2
Points earned:
out of a possible 4 points
Math 234
Exam # 3
TABLE OF L APLACE T RANSFORMATIONS
Page 13 of 14
Math 234
Exam # 3
Page 14 of 14
TABLE OF I NTEGRALS
(1)
'
sin(x) dx = − cos(x) + c
(3)
'
1
sin (x) dx = (x − sin(x) cos(x)) + c
2
(5)
'
sin2 (x) cos(x) dx =
(7)
'
1
cos(x)
+c
dx = −
2
sin(x)
sin (x)
(9)
'
ax
a
ln(1 + bx2 ) + c
dx =
1 + bx2
2b
(11)
'
(13)
'
2
1
sin3 (x) + c
3
(
)
x2 ex dx = x2 − 2x + 2 ex + c
eax sin(bx) dx =
eax
(a sin(bx) − b cos(bx)) + c
a2 + b 2
(2)
'
cos(x) dx = sin(x) + c
(4)
'
cos2 (x) dx =
(6)
'
1
cos2 (x) sin(x) dx = − cos3 (x) + c
3
(8)
'
1
sin(x)
dx =
+c
cos2 (x)
cos(x)
1
(x + sin(x) cos(x)) + c
2
(10)
'
xex dx = (x − 1)ex + c
(12)
'
eax cos(bx) dx =
eax
(a cos(bx) + b sin(bx)) + c
+ b2
a2